Find the sum of possible real values of `x` for which the sixth term of `(3^(log_3 sqrt(9^|x-2|))+7^(1/5 log_7 (3^(|x-2|-9))))^7` equals 567.

Answer

To solve the problem, we need to find the sum of possible real values of \( x \) for which the sixth term of the expression

\[ (3^{\log_3 \sqrt{9^{|x-2|}}} + 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})})^7 \]

equals 567.

### Step-by-Step Solution:

1. **Identify the sixth term**:
The expression is in the form of \( (a + b)^7 \). The sixth term, \( T_6 \), can be expressed as:
\[ T_6 = \binom{7}{5} a^2 b^5 \]
where \( a = 3^{\log_3 \sqrt{9^{|x-2|}}} \) and \( b = 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})} \).

2. **Calculate the binomial coefficient**:
\[ \binom{7}{5} = \frac{7!}{5! \cdot 2!} = \frac{7 \cdot 6}{2 \cdot 1} = 21 \]

3. **Simplify \( a \)**:
\[ a = 3^{\log_3 \sqrt{9^{|x-2|}}} = 3^{\log_3 (9^{|x-2|})^{1/2}} = 3^{\frac{1}{2} \log_3 (9^{|x-2|})} = 3^{\frac{1}{2} \cdot 2|x-2|} = 3^{|x-2|} \]

4. **Simplify \( b \)**:
\[ b = 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})} = 3^{|x-2|-9} \]

5. **Substituting \( a \) and \( b \) into \( T_6 \)**:
\[ T_6 = 21 \cdot (3^{|x-2|})^2 \cdot (3^{|x-2|-9})^5 \]
\[ = 21 \cdot 3^{2|x-2|} \cdot 3^{5(|x-2|-9)} = 21 \cdot 3^{2|x-2| + 5|x-2| - 45} = 21 \cdot 3^{7|x-2| - 45} \]

6. **Set \( T_6 \) equal to 567**:
\[ 21 \cdot 3^{7|x-2| - 45} = 567 \]

7. **Divide both sides by 21**:
\[ 3^{7|x-2| - 45} = \frac{567}{21} = 27 \]

8. **Express 27 as a power of 3**:
\[ 27 = 3^3 \]

9. **Set the exponents equal**:
\[ 7|x-2| - 45 = 3 \]
\[ 7|x-2| = 48 \]
\[ |x-2| = \frac{48}{7} \]

10. **Solve for \( x \)**:
- Case 1: \( x - 2 = \frac{48}{7} \)
\[ x = 2 + \frac{48}{7} = \frac{14 + 48}{7} = \frac{62}{7} \]
- Case 2: \( x - 2 = -\frac{48}{7} \)
\[ x = 2 - \frac{48}{7} = \frac{14 - 48}{7} = -\frac{34}{7} \]

11. **Sum the possible values of \( x \)**:
\[ \text{Sum} = \frac{62}{7} + \left(-\frac{34}{7}\right) = \frac{62 - 34}{7} = \frac{28}{7} = 4 \]

### Final Answer:
The sum of the possible real values of \( x \) is \( \boxed{4} \).